Problem: Let $R$ be the region enclosed by the line $y=3$, the line $x=\dfrac{\pi}{2}$, and the curve $y=3\text{cos}(x)$. $y$ $x$ ${y=3\text{cos}(x)}$ $ R$ $ 0$ $\dfrac{\pi}{2}$ $ 3$ A solid is generated by rotating $R$ about the line $y=3$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi\int^{\frac{\pi}{2}}_0[3-3\text{cos}(x)]^2\, dx$ (Choice B) B $\pi\int^{\frac{\pi}{2}}_0[3-3\text{cos}(x)]\, dx$ (Choice C) C $\pi\int^3_0[3-3\text{cos}(x)]\, dx$ (Choice D) D $\pi\int^3_0[3-3\text{cos}(x)]^2\, dx$
Explanation: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=3\text{cos}(x)}$ Each slice is a cylinder. Let the thickness of each slice be $dx$ and let the radius of the base, as a function of $x$, be $r(x)$. Then, the volume of each slice is $\pi [r(x)]^2\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(x)]^2\,dx$ This is called the disc method. What we now need is to figure out the expression of $r(x)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=3\text{cos}(x)}$ $ 0$ $\dfrac\pi 2$ $ 3$ $r$ The radius is equal to the distance between the curve $y=3\text{cos}(x)$ and the line $y=3$. In other words, for any $x$ -value, this is the equation for $r(x)$ : $\begin{aligned} r(x)}&=3-3\text{cos}(x)} \end{aligned}$ Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(x)}]^2 \\\\ &=\pi\left(3-3\text{cos}(x)}\right)^2 \end{aligned}$ The leftmost endpoint of $R$ is at $x=0$ and the rightmost endpoint is at $x=\dfrac{\pi}{2}$. So the interval of integration is $\left[0,\dfrac{\pi}{2}\right]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int^{\frac{\pi}{2}}_0 \left[\pi(3-3\text{cos}(x))^2\right]\, dx \\\\ &=\pi\int^{\frac{\pi}{2}}_0[3-3\text{cos}(x)]^2\, dx \end{aligned}$